My brute force approach:
We notice that some ranks have more "straight feasibility" than other ranks. An ace has a range of $8$ potential 2nd cards $\{2-5, 10-K\}$, whereas a $2$ has a range of only $5$ potential $2$nd cards $\{A, 3-6\}$.
There are $7$ ranks that have a range of $8: \{A, 5-10\}$. That leaves $6$ ranks left: $2-4$ and $J-K$. We notice again that these $6$ ranks can form $3$ pairs: $2-K$ has a range of $5, 3-Q$ range is $6$, and $4-J$ range is $7$
If $1$st card in $\{A, 5 -> 10\}$, range $r = 8$.
The remaining cards are paired: $$ \begin{align} 2 - K: r &= 5 \\ 3 - Q: r &= 6 \\ 4 - J: r &= 7 \end{align} $$ Applying the Law of Total Probability:
$$P = \frac{7\cdot4}{52} \cdot \frac{8\cdot4}{51} + \frac{8}{52} \cdot \frac{4}{51}(5 + 6 + 7)$$
- $7\cdot4/52$: $P$ (first card in $\{A, 5-10\}$)
- $8\cdot4/51$: Range has $8$ ranks, each rank can have $4$ suits, multiplied together to get the number of $2$nd cards that would satisfy our condition
- $8/52$: $P$ (first card is a "paired" card: $2-4$ or $J-K$)
- $4/51$: potential suits for the rank of the $2$nd card.
- $5, 6, 7$: respective range $r$ of $2$nd card for each pairs
$P = 55.5$%
Is the answer correct? Is there a more elegant / less algebra-extensive approach?
Thank you!